推论2 ds(a< 证：-f(x)≤f(x)≤f(x)川 -fw)ld≤2f)d≤fx)d 即 d ds 性质6设M=maxf(x),m=minf(x),则 [a,b] [a,b] mb-a)≤∫f(x)dr≤M(b-a(a<b) BEIJING UNIVERSITY OF POSTS AND TELECOMMUNICATIONS PRESS 录 返回 结束目录 上页 下页 返回 结束 推论2 f x x b a ( )d  f x x b a ( ) d   证:   f (x)  f (x)  f (x) (a  b) f x x f x x f x x b a b a b a ( ) d ( )d ( ) d        即 f x x f x x b a b a ( )d ( ) d    性质6 设 max ( ), min ( ) , [ , ] [ , ] M f x m f x a b a b   则 m(b a) f (x)dx M (b a) b a      (a  b)